Answer:
8.800s
Explanation:
When the performer swings, she oscillates in SHM about Lo of the string with time period To = 8.90s.
First, determine the original length Lo, where for a SHM the time period is related to length and the gravitational acceleration by the equation
T = 2π×√(Lo/g)..... (1)
Let's make Lo the subject of the formulae
Lo = gTo^2/4π^2 ..... (2)
Let's put our values into equation (2) to get Lo
Lo = gTo^2/4π^2
= (9.8m/s^2)(8.90s)^2
------------------------------
4π^2
= 19.663m
Second instant, when she rise by 44.0cm, so the length Lo will be reduced by 44.0cm and the final length will be
L = Lo - (0.44m)
= 19.663m - 0.44m
= 19.223m
Now let use the value of L into equation (1) to get the period T after raising
T = 2π×√(L/g)
= 2π×√(19.223m/9.8m/s^2)
= 8.800s
The part of the phospholipid molecule that will face the water, as it is labelled to be hydrophilic would be the polar phosphate group. The correct response would be A.
Answer: Your answer is True.
Explanation:
Answer:
The independent variable is amounts of distilled water.
Explanation:
The dependent variable of this experiment is the height of each plant which is dependent on the amounts of distilled water added to each pot. Notice that days and amount of soil are constants. Therefore, the independent variable in this case is the amount of water added to each pot.
